- Sections:
- 0, Chapter 1, Introduction,
### Introduction

The following units of work on calculus bring together extended methods in both Differentiation and Integration. After studying this unit you should

- be able to differentiate and integrate basic trigonometric functions
- understand how to calculate rates of change
- be able to integrate using substitutions
- be able to integrate by parts
- be able to formulate and solve simple first order differential equations.

The topic is divided into the following 5 sections:

- Derivatives of Trigonometric Functions
- Rates of Change
- Integration by Substitution
- Integration by Parts
- First Order Differential Equations

- 0, Chapter 2, Derivative of Trigonometric Functions,
### Derivative of Trigonometric Functions

We start by reminding ourselves of the definition of a derivative of a general function

The derivative of a general function \(f\left(x\right)\) is given by \(\frac{df}{dx} = ^{\lim }_{h \rightarrow 0} \left\{\frac{f\left(x+h\right) -f\left(h\right) }{h} \right\}\)

You are going to obtain the derivatives of \(\sin x\), \(\cos x\), and \(\tan x\).

Remember that the derivative of a function at a given point is given by the gradient of that function at the given point.

Suppose you look at a sketch of the graph \(\sin x\) and sketch in the tangents for values of \(x\)at the points below

\( -2 \pi\), \(-\frac {3 \pi}{2}\), \( -\pi\), \(-\frac{\pi}{2}\), \(0\), \(\frac{\pi}{2}\), \(\pi\), \(\frac{3 \pi}{2}\), \(2 \pi\)This is shown below.

These tangent values when plotted against \(x\) will look like the sketch below.

This suggests that the graph of the function for the tangent may be something like the sketch opposite which suggests that the derivative of \(\sin x\) might be \(\cos x\), but this illustration is certainly not a proof.

- 1, Chapter 2.1, Geometric proof,
### Geometric proof

It is also possible to use a circle of unit radius to show that, for example

\(\frac{df}{dx} = \left(\sin x\right) = \cos x\)In the diagram shown below \(h\) is small, and

\(A′ B′ = \sin \left(x + h\right)\) and \(AB = sin x\)

Hence \(\sin \left(x + h\right) – sin x = A′ P\), and

\(A′ A = radius \times angle = h\)Now assume that when \(h\) is very small the arc \(A′ A\) is approximately straight.

For the triangle \(A A′ P\),we can deduce\(\frac {\sin \left(x + h\right) - \sin x}{h} = \frac {A′ P}{A′ A} \approx \cos x\)

So, approximately, if \(h\) is very small

\( \frac{\sin \left(x+h\right) - \sin x}{h} \approx \cos x\)And taking the limit as \( h \rightarrow 0\), \(\frac {d}{dx} \left(\sin x\right) = \cos x\)

- 1, Chapter 2.2, Key Results,
### Key Results

Below are the three crucial derivatives and their inverse integration:

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\(\frac{d}{dx}\left(\sin x\right) = \cos x\) \(\frac{d}{dx}\left(\cos x\right) = - \sin x\) \(\frac{d}{dx}\left(\tan x\right) = - \sec^{2} x\)

\(\int{\cos x\;dx = \sin x+c } \) \(\int{\sin x\;dx = -\cos x+c } \) \(\int{\sec^{2} x\;dx = \tan x+c }\)

**Revision of secant, cotangent and cosecant.**\(\sec \theta = \frac{Hypotenuse}{Adjacent} = \frac{1}{\cos \theta}\)

\(\cot \theta = \frac{Adjacent}{Opposite} = \frac{1}{\tan \theta} = \frac {\cos \theta}{ \sin \theta}\)

\(\csc \theta = \frac {Hypotenuse}{Opposite} = \frac{1}{\sin \theta}\) - 1, Chapter 2.2, Task 1, Worked Example 1,
### Worked Example

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- 1, Chapter 2.3, Trig Revision and Formulae,
### Trigonometric Revision and Formulae

**Key Formulae**Note that \(\sin^{2}x\) means find \(\sin x\), then square the result.

And \(\sin \left(x^{2}\right)\) means to square \(x\), then the sine function.**Trig Formulae**As \(\sin^{2}x + \cos^{2}x =1\) we can deduce

\(\sin^{2} x = 1 - \cos ^{2} x\)

\(\cos ^{2}x = 1 - \sin ^{2} x\)

\(\tan^{2}x = \sec^{2}x-1\)

\(\cot^{2}x= \csc^{2} x -1\)Note - throughtout the unit we write cosec\(x\) as csc\(x\)

**Double Angle Formulae**

\(\sin 2x= 2 \sin x \cos x\)

\(\cos 2x = 1 - 2 \sin^{2} x\)

\(\tan 2x = \frac{2 \tan x}{1 - \tan^{2} x}\)**Differentiating**

\(\frac{d}{dx} \left(\frac{1}{\tan x} \right) = \frac{1}{\left(1 + x^{2}\right) } \)

\(\frac{d}{dx} \left(\frac{1}{\sin x} \right) = \frac{1}{\left(\sqrt{\left(1 -x^{2}\right) } \right) } \)\(\frac{d}{dx} \left(\csc x\right) = - \csc x \cot x\)

\(\frac{d}{dx} \left(\sec x\right) = \sec x \tan x\)

\(\frac{d}{dx} \left(\cot x\right) = - \csc^{2} x\) - 0, Chapter 3, Rates of Change,
### Rates of Change

From your earlier work on differentiation, you will recognise that the rate of change of a variable, say \(f\), is given by \(\frac{df}{dt}\), where \(t\) denotes time. Often though, \(f\) is a function of another variable, say \(x\), so that you need to use the ‘function of a function’ rule,

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Function of Function Rule

\(\frac{df}{dt} = \frac{df}{dx} \frac{dx}{dt}\)

- 1, Chapter 3, Task 1, Worked Example 2,
### Worked Example

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- 0, Chapter 4, Integration by Substitution,
### Integration by Substitution

This method is the equivalent to using a change of variable when differentiating composite functions. In general'

If \(y = \int{f \left(x\right)} dx\)

then \(\frac{dy}{dx} = f \left(x\right)\)We will use this fact to derive a rule for integrating called integration by substitution.

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If we let \(x = g\left(u\right)\) for some function \(g\) of a new variable \(u\), then \(y\) becomes a function of \(u\) and

\(\frac{dy}{du} = \frac{dy}{dx} \times \frac{dx}{du}\)\(\frac{dy}{du} = f \left(x\right) \frac{dx}{du} = f \left(g\left(u\right) \right) \frac{dx}{du} \)

Then

\(y = \int{f \left(g\left(u\right) \right) \frac{dx}{du}} du\). - 1, Chapter 4, Task 1, Worked Example 3,
### Worked Example

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- 0, Chapter 5, Integration by Parts,
### Integration by Parts 1

Our second method of integration is a very important one which is particularly useful for a number of different integrals.

You can see how it works by first differentiating with respect to \(x\), \(\frac{d}{dx} \left(x \sin x\right) = x + \cos x\).Let's explore an example problem first which illustrates the method

By assuming \(\int{xe^{x}dx } =(ax+b)e^{x} +c\) find the values of the constants \(a\) and \(b\) so that the result is valid.

**Solution**We differentiate both sides of the equation above to give:

\(x e^{x}= \frac{d}{dx}((ax+b)e^{x}+c)\)

\(= (ax+b)e^{x}ae^{x}+0\)

\(=(ax+a+b)e^{x}\)So comparing coefficients, \(a=1\)and \(a+b=0\) giving \(b=-1\).

Hence \(\int{xe^{x}dx } =(x-1)e^{x} +c\)

- 1, Chapter 5.1, Integration by Parts 2,
### Integration by Parts 2

By now you should be wondering if there is a more precise method of evaluating integrals of the type above, where the method requires inspired guesswork to start with.

In fact, there is a more formal methodINSERT A REMEMBER BOX HERE

If \(u\) and \(v\) are two functions of \(x\) then from your work on differentiation, you know that:

\(\frac{d}{dx} \left(uv\right) = v \frac{du}{dx} + u\frac{dv}{dx}\)

If we take the formula above and integrate each side with respect to \(x\), we get

\(uv =\int{v\frac{du}{dx} } dx + \int{u\frac{dv}{dx} } dx\) - 1, Chapter 5.1, Task 1, Worked Example 4,
### Worked Example

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- 0, Chapter 6, First Order Differential Equations,
### First Order Differential Equations

You might well wonder why you have studied integration in such depth. The main reason is the crucial role it plays in solving what are known as

**differential equations**.

Here you will only be dealing with first order differential equations, which can be expressed, in general, in the form\(\frac {dy}{dx} = f \left(x,y\right)\)

when \(y = f \left(x,y\right)\) is an unknown function of \(x\), and \(f\) is a given function of \(x\) and \(y\).

Examples of such equations include

\(\frac {dy}{dx} = xy\)

\(\frac {dy}{dx} = x + y\)

\(\frac {dy}{dx} = \frac {x}{y}\)

\(\left( y \neq 0\right)\)

We will explore solving such equations by demonstration in the worked examples that follow

- 1, Chapter 6, Task 1, Worked Example 5,
### Worked Example

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- 0, Chapter 7, Summary,
### Summary

**Differentiation**of trig functions:\(\frac{d}{dx} \left(\sin x\right) =\cos x\)

\(\frac{d}{dx} \left(\cos x\right) = - \sin x\)

\(\frac{d}{dx} \left(\tan x\right) = \sec^{2} x\)

\(\frac{d}{dx} \left(\cot x\right) = - \csc^{2} x\)**Integration**of trig functions\(\int{\cos x\;dx = \sin x+ c }\)

\(\int{\sin dx = - \cos x+ c } \)

\(\int{\sec^{2} dx = \tan x+ c } \)

\(\int{\csc^{2} x\;dx = - \cos x+ c } \)**Function of a function**: if \(y = f \left(x\right)\) and \(x = g \left(t\right)\), then

\(\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\)**Integration by substitution**: if \(y=\int{f\left(x\right) } dx\) and \(x = g \left(u\right) \), then

\(y = \int{f\left(g\left(u\right) \right) } \frac{du}{dx} dx\)**Integration by parts**: if \(n\) and \(v\) are functions of \(x\), then

\(\int{u\frac{dv}{dx} } dx = uv - \int{v\frac{du}{dx} } dx\)A

**first order differential equation**is of the form

\(\frac{dy}{dx} = f \left(x,y\right)\)

where \(f\) is a function of \(x\) and \(y\).

- 0, Chapter 1, Introduction,
- Interactive Exercises:
- Even More Calculus , https://www.cimt.org.uk/sif/calculus/c5/index.htm
- Text Exercises, https://www.cimt.org.uk/sif/calculus/c5/text.pdf
- Answers, https://www.cimt.org.uk/sif/calculus/c5/text_ans.pdf

- YouTube URL: https://www.youtube.com/watch?v=G0LL8jeO0t4&list=PL8ce1n6mRlCtS0rO_C7CNwtjhyqDVkr8g&index=48

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